Maths Quiz on Differential Equations for IIT JEE & Engineering Exam

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Maths Quiz on Differential Equations
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JEE Maths Quiz on Differential Equations

JEE Maths Quiz on Differential Equations : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you quiz on Differential Equations.

Quiz on Differential Equations

 

Q1.  If x dy = y (dx + y dy), y > 0 and y (1) = 1, then y (−3) is equal to _________.
a) 0
b) 3
c) 5 
d) -3

View Answer

b) 3

Solution:

x dy = y (dx + y dy)

[x dy − ydx] / [y2] = dy

−d (x / y) = dy

Integrating both sides, we get x / y + y = c [Because y (1) = 1 ⇒ c = 2; Hence xy + y = 2 ]

For x = −3,

y− 2y − 3 = 0

⇒ y = −1 or 3

⇒y = 3 (Because y > 0)

Q2.  (x+ y2) dy = xy dx. If y (x0) = e, y (1) = 1, then the value of x0 = __________.
a) ± √2e
b) ± √3e
c) ± √4e
d) ± √5e

View Answer

b) ± √3e

Solution:

(x+ y2) dy = xy dx

x (x dy − y dx) = −y2 dy

[x * (y dx − x dy)] / y2 = dy

[x / y] d (x / y) = dy / y

Integrating, x/ 2y2 = loge y + c

Given y (1) = 1

c = 1 / 2

x/ 2y= loge y + 1 / 2

Now y (x0) = e

x0/ 2e− loge e − 1 / 2 = 0

x0= 3e2

x= ± √3e

Q3. Solution of differential equation 2xy dy / dx = x2 + 3y2 is __________.
a) x2 + y= px2
b) x2 – y= px3
c) x3 + y3 = px2
d) x2 + y= px3

View Answer

d) x2 + y= px3

It is homogeneous equation dy / dx = [x2 + 3y2] / 2xy

Put y = vx and dy / dx = v + x * [dv / dx]

So, we get x * [dv / dx] = [1 + v2] / 2v

2v dv / 1 + v2 = dx / x

On integrating, we get

x2 + y= px3. (where p is a constant)

Q4.  An integrating factor for the differential equation (1 + y2) dx − (tan−1 y − x) dy = 0
a) etan−1y
b) etan+1y
c) etan−2y
d) etan+2y

View Answer

a) etan−1y

Solution:

(1 + y2) dx − (tan−1 y − x) dy = 0

dy / dx = (1 + y2) / (tan−1 y − x)

dx / dy = [tan−1 y / 1 + y2] − [x / 1 + y2] [dx / dy] + [x / 1 + y2] = [tan−1 y] / [1 + y2]

This is equation of the form dx / dy + Px = Q

So, I.F. = e∫Pdy

= e∫1 / 1 + y2.dy

= etan−1y

Q5. What is the general solution of the differential equation (2x − y + 1) dx + (2y − x + 1) dy = 0?

View Answer

Solution:

(2x −y + 1) dx + (2y − x + 1) dy = 0

dy / dx = 2x − y + 1x − 2y −1, put x = X + h, y = Y + k

dY / dX = [2X − Y + 2h − k + 1] / [X − 2Y + h −2k − 1]

2h − k + 1 = 0

h − 2k − 1 = 0

On solving h = −1, k = −1;

dY/ dX = [2X − Y] / [X − 2Y]

Put Y = vX;

dY / dX = v + [X dv / dX]

v + [X dv / dX] = [2X − vX] / [X − 2vX] = [2 − v] / [1 − 2v]

X dv / dX = [2 − 2v + 2v2] / [1 − 2v] = 2 (v2 −v + 1) / [1 − 2v]

dX / X = (1 − 2v) / 2 (v− v + 1) dv

Put v− v + 1 = t

(2v − 1) dv = dt

dX / X = −dt / 2t

log X = logt− ½ + log c

X = t−1/2 c

X = (v2 − v + 1)−1/2 * c

X(v2 − v + 1) = constant

(x + 1)(([y + 1)/ (x + 1)2] − [(y+1) / (x + 1)] + 1) = constant

(y + 1)− (y + 1) (x + 1) + (x + 1)= c

y2 + x2 − xy + x + y = c

Q6.  If y′= [x − y] / [x + y], then its solution is ____________.

View Answer

Solution:

Given dy / dx = x −y / x + y.

Put y = vx

dy / dx = v + x * [dv / dx]

v + x * [dv / dx] = [x − vx] / [x + vx]

v + x [dv / dx] = [1 − v] / [1 + v] [1 + v] / [2 − (1 + v)2] dv = dx / x

Integrating both sides, ∫[1 + v] / [2 − (1 + v)2] dv = ∫dx / x

Put (1 + v)= t

⇒ 2 (1 + v) dv = dt

[1 / 2] ∫dt / [2 − t] =∫dx / x

[− 1 / 2] log (2 − t) = log xc

[−1 / 2] log [2 − (1 + v)2] = log xc

[−1 / 2] log [−v− 2v + 1] = log xc

log 1 / √[1 − 2v −v2] = log xc

x2c(1 − 2v −v2) = 1

y2 + 2xy − x2 = c1.

Q7. A function y = f (x) has the second-order derivatives f′′(x) = 6 (x − 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x − 5, then the function is __________.
a) (x + 1)3
b) (x − 1)2
c) (x − 1)3
d) (x + 1)2

View Answer

c) (x − 1)3

Solution:

Given f′′(x) = 6 (x − 1)

f′(x) = 3 (x − 1)+ c1 ..(i)

But at point (2, 1) the line y = 3x − 5 is tangent to the graph y = f(x).

Hence, dy / dx∣x = 2 = 3 or

f′ (2) = 3.

Then from (i) f′ (2) = 3 (2 − 1)+ c1

3 = 3 + c1

c= 0 i.e.,

f′ (x) = 3 (x − 1)2

Given f (2) = 1

f (x) = (x − 1)+ c2

f (2) = 1 + c2

1 = 1 + c2

c= 0

Hence, f (x) = (x − 1)3.

Q8. The general solution of the differential equation (x + y) dx + x dy = 0 is _______.

View Answer

Solution:

(x + y) dx + x dy = 0

x dy = − (x + y) dx

dy / dx = −[x + y] / x

It is homogeneous equation, hence, put y = vx and dy / dx = v + x [dv / dx], we get

v + x dv / dx = −[x + vx] / x= −[1 + v] / 1

x dv / dx = −1 −2v

∫dv / [1 + 2v] = −∫dx / x

[1 / 2] log (1 + 2v) = −log x + log c

log (1 + 2 [y / x]) = 2 log [c / x] [x + 2y] / x = (c / x)2

x+ 2xy = c.

Q9. Equation of curve through the point (1, 0) which satisfies the differential equation (1 + y2) dx − xy dy = 0, is ________.
a) 3
b) 5
c) 2 
d) 1

View Answer

d) 1

We have [dx / x] = [y dy] / [1 + y2]

Integrating, we get log |x| = ([1 / 2] * log [1 + y2]) + log c

|x| = c √(1 + y2)−−−−−−−

But it passes through (1, 0), so we get c = 1

Therefore, the solution is x2 = y2 + 1 or x2 − y= 1.

Q10. The equation of the curve which passes through the point (1, 1) and whose slope is given by 2y / x, is __________.
a) x2
b) 2x2
c) 1/2x2
d) x3

View Answer

 a) x2

Slope dy / dx = 2y / x

2 ∫dx / x =∫dy / y

2 log x = log y + log c

x= yc

Since it passes through (1, 1), therefore, c = 1.

Hence, x− y = 0

y = x2.

 



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